Two small globules AB with masses of m1 m2 with equal amounts of xenocharge

Answer: The mechanical energy of the DA system keeps increasing ......Wrong, the whole process of the two springs do a cyclic motion of contraction and expansion, similar to a spring oscillator.

b Conservation of mechanical energy of the system ......Error, the rest starts, the electric field force does positive work on both a and b, and the mechanical energy increases; The mechanical energy of the return process is reduced again.

c The system momentum is increasing ......False, the electric field force is to the left and to the right to b, because the electric charge is equal, so the electric field force is equal, so the momentum is conserved.

d The system is conserved in momentum ......That's right.

Ball A receives an electric field force to the left, and ball B receives an electric field force to the right, and the two electric field forces are equal in magnitude. During the movement to the left, the spring elastic force increases, the acceleration decreases, and when the equilibrium position is reached, the velocity is maximum; Continue to move to the left, the spring tension continues to increase, the acceleration to the right, that is, to do deceleration motion, when the speed is 0, the acceleration is the maximum, return to the motion; After that, the process is repeated repeatedly, and it is actually a simple harmonic vibration process. b Same thing.

Conditions for the application of the law of conservation of momentum (1) The system is not subjected to an external force or the net force of the external force to which the system is subjected is zero.

2) Although the net force of the external force of the system is not zero, it is much smaller than the internal force of the system.

3) Although the net force of the external force on the system is not zero, but the component force in a certain direction is zero, then the total momentum of the system remains unchanged in that direction - the partial momentum is conserved.

4) In some practical problems, the sum of the external forces on a system is not zero, and the internal forces are not much greater than the external forces, but the projection of the external forces in a certain direction is zero, then the condition of conservation of momentum in this direction is also satisfied.

The ball AB will move all the way to the left, and the tension of the spring will increase more and more until the spring breaks, which is normal. The question gives the spring within the elastic limit, so how to stipulate in the second article.

After the Cd is electrified, A will move to the left and B will move to the right, the spring will be elongated, and both balls will accelerate before the spring force is less than the electric field force; When the elastic force is equal to the electric field force, the velocity of the two balls reaches the maximum, that is, the kinetic energy is the maximum; As the spring continues to lengthen the orange, the elastic force exceeds the electric field force, the two balls decelerate the motion, the direction of motion remains unchanged, when the speed is reduced to 0, the spring length reaches the maximum value, in this process, the electric field has been doing positive work, so the mechanical energy of the system is the largest at this time.

1) The left ball is positively charged, and the right ball is negatively charged. (Judging by the electric field force and the direction of the electric field) when the external electric field is not added, the two balls are balanced, and each ball is balanced under the action of three forces: gravity, Coulomb force, and rope tension.

Find the geometric relationship and draw a force equilibrium triangle with the distance between the two balls L [2l-l 2-l 2] to get the Coulomb force gravity = tan30° kq l = mgtan30°

q=√(mgtan30°/k) l

2) The ball is balanced under the action of 4 forces: gravity, Coulomb force, electric field force, rope tension force, and the distance between the two balls is 3L [2L+L 2+L 2].

Horizontal resultant force = electric field force - Coulomb force.

Horizontal resultant force gravity = tan30°

qe-kq (3L) =mgtan30° [kq l =mgtan30° in the first question].

qe-mgtan30°/9=mgtan30°e=10mgtan30°/9q

Although there is no figure to determine that ma must be less than mb a, the reason is that the two spheres are equal in magnitude by the Coulomb force, and the magnitude is f=mgtanq

f=kq1q2 r 2 cannot determine the amount of power qa must be greater than qb error.

mgl(1-cosq)=1 2mv 2 va must be better than vb pairs.

The maximum kinetic energy is equal to mgh(1-cos) cos = mghtan tan(2), since mghtan is equal, the maximum kinetic energy of the starting big ball is greater. (tan( 2)=(1-cos) sin ) Add h is the height of the spherical rock to the overhang, which should be clear.

1) The left ball is positively charged, and the right ball is negatively charged. (Judging by the electric field force and the direction of the electric field) when the external electric field is not added, the two balls are balanced, and each ball is small.

It's all in gravity. The Coulomb force rope tension is balanced under the action of 3 forces.

Looking for the geometric relationship, the distance between the two balls is l

2l-l/2-l/2】

Draw a force balance triangle.

The Coulomb force is obtained with gravity = tan30°

kq²/l²=mgtan30°

q=√(mgtan30°/k)

l2) The ball is in.

Gravity Coulomb force.

The electric field force rope tension is balanced under the action of 4 forces, and the distance between the two balls is 3L [2L + L 2 + L 2].

Horizontal resultant force = electric field force - Coulomb force.

Horizontal resultant force gravity = tan30°

QE-KQ (3L) = MGTan30° [in the first question.

kq²/l²=mgtan30°

qe-mgtan30°/9=mgtan30°e=10mgtan30°/9q

Option D shouldn't be right, I know there are a lot of answers out there that are ACD.

Under the premise that the mass of ball A is less than the mass of ball B, the gravitational potential energy of B is greater than a at the same altitude, and the gravitational potential energy of A is less than that of B (relative to the ground) at the lowest point, and it is impossible to know which kinetic energy is larger.

From another point of view, although the height difference between A and B is larger than that of B, it is impossible to know which of the kinetic energy converted by gravitational potential energy is greater or smaller because the mass of A is smaller than that of B.

Thinking about it from another angle, since the height difference of A is larger than that of B, then it means that the velocity of A is greater than that of B (the speed of falling has nothing to do with mass), and the mass of A is smaller than that of B, so the kinetic energy at the lowest point is still not comparable.

Think about it again, and the answer copied and pasted by the brother LS finally deduced ek=mgtan, I didn't derive it, for the time being, when his answer was correct, and then he said, because the thin line connected by A and the vertical plane angle are large, so the tan value is large, so the ek is large. But he also ignores the fact that the mass of A is smaller than that of B, so even if the Tan value is larger, the EK relation cannot be deduced at all because the mass of A itself is smaller than that of B.

In other words, if the quality of A is less than that of B, the answer to D cannot be selected at all. I don't know if my understanding is correct, if it's wrong, everyone comments and corrects it, thank you. If it is correct, it will be miserable, the answers to the college entrance examination questions are wrong, and the candidates will be wronged.

My answer is AC, if the teacher says it is wrong, please explain it, otherwise, this question will be invalid.

Conservation of mechanical energy:

ek=mg△h=f/tanθ ×h(1-cosθ)/cosθ=fh(1-cosθ)/sinθ

In this equation, for balls a and b, the electrostatic force f is of the same magnitude, but is different.

After solving this step, I personally think that it is easier to take two special values (e.g. 30 degrees and 45 degrees) than to solve trigonometric functions, and this is also a common way to solve multiple-choice problems in mathematics and physics (take special value method).

That's right on the first floor... I won't stop screaming on the second floor, begging you! m is not the same, resulting in different angles...

Analysis: (1) From the situation in Figure B, it can be seen that ball A is positively charged and ball B is negatively charged.

In the absence of a homogenized strong electric field, the situation is as shown in Figure A, at this time, for ball A, it is subjected to gravity mg (straight down), ball B to its gravitational force f1 (horizontal to the right), rope pull force t1 (along the rope upward), the net force is 0.

f1 mg) tan

f1＝k* q^2 / 2l－2* lsinα)^2

The absolute value of the charge carried by each ball is q 2l*(1 sin)*root number (mg *tan k).

Q 2L * (1 sin30 degrees) * root number (mg * tan30 degrees k) l * root number.

2) When the uniform electric field to the left is added, the situation is as shown in Figure B, at this time, for ball A, it is subjected to gravity mg (straight down), the gravitational force of ball B on it f2 (horizontally to the right), the rope pull force t2 (along the rope upward), and the force of the uniform electric field on it f3 (horizontally to the left), and the net force is 0.

At this time f2 k* q 2 l 2 * lsin ) 2

According to the angle between the rope and the vertical direction is still equal to 30 degrees, we can know f3 f2 f1

i.e. q e [k* q 2 2l 2* lsin ) 2 ] k* q 2 2l 2* lsin ) 2

Substituting 30 degrees, q l *root number into the above equation and simplifying it, we get.

e＝ /9 l )

(1) The ball A is positively charged and B is negatively charged.

by f=mgtan30 0=kq 2 l 2q=(mgtan30 0l 2 k) 1 2(2) by equilibrium condition.

qe=kq^2/9l^2+mgtan30^0e=(kmgtan30^0/81l^2)^1/2+(kmgtan30^0/l^2)^1/2

<> equation is listed, and the bridge is fierce, so let's solve it by clearing the ruler.